JEE Advanced · Chemistry · 6. Thermodynamics (C)
One mole of an ideal gas is taken from \(a\) to \(b\) along two paths denoted by the solid and the dashed lines as shown in the graph below. If the work done along the solid line path is \(W_s\) and that along the dotted line path is \(W_d\), then the integer closest to the ratio \(\frac{W_d}{W_s}\) is
- A 1
- B 2
- C 3
- D 4
Answer & Solution
Correct Answer
(B) 2
Step-by-step Solution
Detailed explanation
Solid line represents reversible isothermal process.
So, work \(W_s=-4 \times 0.5 \ln \left(\frac{5.5}{0.5}\right)=-2 \ln 11 \mathrm{~L}-\mathrm{atm}\)
Dotted line represents irreversible process
So, work
\(
W_d=-\left\{4 \times 1.5+1 \times 1+\frac{2}{3} \times 2.5\right\}=\) \(-\left\{6+1+\frac{5}{3}\right\} \mathrm{L}-\mathrm{atm}=-\frac{26}{3} \mathrm{~L}-\mathrm{atm}
\)
So \(\quad \frac{W_d}{W_s}=\frac{26}{3 \times 2 \ln 11} \approx 2\)
Energetics
Conceptual
III
So, work \(W_s=-4 \times 0.5 \ln \left(\frac{5.5}{0.5}\right)=-2 \ln 11 \mathrm{~L}-\mathrm{atm}\)
Dotted line represents irreversible process
So, work
\(
W_d=-\left\{4 \times 1.5+1 \times 1+\frac{2}{3} \times 2.5\right\}=\) \(-\left\{6+1+\frac{5}{3}\right\} \mathrm{L}-\mathrm{atm}=-\frac{26}{3} \mathrm{~L}-\mathrm{atm}
\)
So \(\quad \frac{W_d}{W_s}=\frac{26}{3 \times 2 \ln 11} \approx 2\)
Energetics
Conceptual
III
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